Multiples of determinant are elements in a matrix Suppose we have an $n \times n$ square matrix, $A$. Let the determinant be $|A|.$ We also constrain the elements of $A$ such that each element of  $A$ is an integer multiple of $|A|.$ Is there an equation to generate such a matrix?
 A: Clearly, the zero matrix of any size is such an $A$. On the other hand, a non-zero matrix with zero determinant is not such an $A$. Going forward, we take $A$ to be a (square) non-zero integer matrix with non-zero determinant.

Suppose that multiplying some (integer) matrix $M$ by $|A|\neq 0$ gives $A$, and recall that multiplying the entries of a single row of a matrix by some value multiplies the determinant by that value. Thus, for $n$-row matrices $M$ and $A$, we have
$$A = |A|\; M \quad \implies \quad |A| = |A|^n |M| \quad \implies \quad |M| = \frac{1}{|A|^{n-1}}$$
Since $|A|$ is an integer, and since $|M|$ must be an integer as well, we can conclude that $|A|^{n-1}$ must be $\pm 1$, whereupon $n=1$ (and $A$ is any non-zero $1\times 1$ matrix), or else $|A|$ itself is $\pm 1$. We can see that these conditions are also sufficient.

So, the possible solutions are


*

*The zero matrix of any size.

*Any $1\times 1$ matrix.

*Any matrix with determinant $\pm 1$.

A: Basically you have the following equation $$\det(A)=\left(\det(A)\right)^n \det(B)$$ where $$A=\det(A)B$$ If $B$ is invertible, then you can get $$\det(A)=\left(\det(B)\right)^{-1/(n-1)}$$ So, if you find any integer invertible matrix $B$, then you can construct $A$ by multiplying each element of $B$ by $\det(A)$ as obtained above. But $\det(A)$ does not seem to turn out to be an integer in general.
A: The identity is such a matrix, as is any other integer matrix of determinant $\pm 1$. So is the zero matrix.
